I am a beginner programmer and there was this exercise I found that said.

Write a string of characters and determine the number of words, numbers, uppercase and lowercase characters and spaces. I thought what I build was a decent enough program and it works, kinda! The problem is that when I try to run it the result is not entirely correct. For example; When I write: HI MY name is Ani 1 1 2 a it says that

• Spaces=8 correct here

• Numbers= 3 correct here as well

• Upper Case characters=4 It should be 5

• Lower Case characters = 7 it should be 9

• Words = 26 which is completely wrong

As for the words, I found a new way to count them. By counting spaces 1 but I want to count them correctly

Is it possible to point out the mistakes?

This is what I have done so far

int main() {
    char str[1000 1];
    int words = 0;
    int numbers = 0;
    int uppercharacters = 0;
    int lowercharacters = 0;
    int spaces = 0;
    int i;

    printf("Please enter the string of characters: ");
    gets(str);
    for (i = 0; str[i] != '\0'; i  ) {
        if (str[i] > 'a' && str[i] < 'z')
            lowercharacters  ;
        else if (str[i] > 'A' && str[i] < 'Z')
            uppercharacters  ;
        else if (str[i] == ' ')
            spaces  ;
        else if (str[i] > '0' && str[i] < '9')
            numbers  ;
        else if (str[i] == ' ' && str[i   1] != ' ');
            words  ;
    }
    printf("Spaces = %d\n", spaces);
    printf("numbers = %d\n", numbers);
    printf("Upper Case characters = %d\n", uppercharacters);
    printf("Lower Case characters = %d\n", lowercharacters);
    printf("Words = %d\n", words   1);
    return 0;
}

CodePudding user response：

As for the words i found a new way to count them. By counting spaces 1 but i want to count them correctly

Code fails due to ; at the end of the else if().
Tip Good compilers with all warnings enabled will warn about that.
Save time, enable all warnings.

//                                   v !!! 
else if(str[i]==' ' && str[i 1]!=' ');
    words  ;

Even if corrected to

else if(str[i]==' ' && str[i 1]!=' ') 
    words  ;

Still fails with input like " abc" (lead space) reports as 2 words

Instead count the occurrences of a letter following a non-letter.

char previous = '\n';
 
for(i=0; str[i] != '\0'; i  ) {
  if (isalpha(str[i]) && !isalpha(previous)) {
    words  ;
  }
  previous = str[i]; 
}

Make you own helper functions if standard ones like is...() are not allowed.

CodePudding user response：

You should use if(str[i]>='a' && str[i]<='z') instead of if(str[i]>'a' && str[i]<'z'). You don't want to exclude the characters z and a from being tested.

CodePudding user response：

For the counting words part, notice that there is one misplaced semicolon after your last else if statement. The number of words won't be 100% correct if you fix that typo, but you might be able to work from there :)